Consider this as a function of (note the negative sign in front of the term in
I presume is a mistake), the derivative of this is:

which is a geometric series and it converges for .

So write down the sum of this last series, then integrate with the constant
of integration set so that the integral is zero for , and that
is your answer (it looks like (d) to me but you will need to check).

RonL

May 22nd 2007, 03:28 PM

ThePerfectHacker

There are many verions, I use the the following one.

Taylor's Theorem (Lagrange): Let be differenciable on (with ) and for the remainder is given by for some between and . Where the remainder is the difference between the function and its degree Taylor polynomial , defined as and .